Proof Let $n\in \omega $ such that $n\in \omega \setminus dom\left ( E_{q}\right ) $ and $f:A_{p}\cap n\longrightarrow 2,$ we need to prove that $\overline {f}$ is well-defined. Let $b\in A_{q}\cap n,$ we need to argue that $\overline {f}( b) $ is actually defined. We proceed by cases:

Here, $\overline {f}( b) $ is defined by one of the first three clauses in the definition of $\overline {f}$ (if b is a leader in $E_{p},$ then the first clause applies, otherwise, either clause 2 or 3 applies).

Here, we have that $\overline {f}( b) =H_{p}^{b}\left ( f\upharpoonright \left ( A_{p}\cap b\right ) \right ) $ and this is well-defined by point 4 of the definition of $p\leq q.$ ▪

Proof We will prove that $\dot {x}_{gen}$ is forced to intersect every element of $\mathcal {U},$ the proof for $\omega \backslash \dot {x}_{gen}$ is similar. Let $B\in \mathcal {U}, p\in \mathbb {S}( \mathcal {U}) $ and $n\in \omega .$ It is enough to prove that there is $q\leq p$ such that $q\Vdash \text{``}B\cap \dot {x}_{gen}\nsubseteq n.\text{''}$ Since $dom( E_{p}) \in \mathcal {U},$ there is $m>n$ such that $m\in B\cap dom( E_{p}) .$ Since $m\in dom( E_{p}) ,$ we know there is a (unique) $a\in A_{p}$ such that $m\in [ a] _{p}.$ Define a condition q with the following properties:

1. 1. $q\leq p.$

2. 2. $dom\left ( E_{q}\right ) =dom( E_{p}) \setminus [ a] _{p}.$

3. 3. If $b\in dom( E_{p}) \setminus [ a] _{p}$ then $[ b] _{p}=[ b] _{q}.$

4. 4. If $m\in Lov_{p}( a) $ then $H_{q}^{a}$ is the constant $1$ function, and if $m\in Hat_{p}( a) ,$ then $H_{q}^{a}$ is the constant $0$ function.

It is easy to see that $q\Vdash \text{``}m\in B\cap \dot {x}_{gen},\text{''}$ so we are done.▪

Proof We first find $q_{1}\leq _{n}^{\ast }q$ such that $q_{1}\in D_{q}.$ Let $S=\left(dom\left ( E_{q}\right ) \backslash dom\left ( E_{q_{1}}\right ) \right)\cup \left(\left [ a_{q_{1}}( n) \right ] _{q_{1}}\backslash \left [ a_{q}( n) \right ] _{q}\right)$ and note it is an element of $\mathcal {U}^{\ast }.$ Let $r_{1}$ be any condition such that:

1. 1. $r_{1}\leq _{n}^{\ast }r.$

2. 2. $\left [ a_{r_{1}}( n) \right ] _{r_{1}}=\left [ a_{r}( n) \right ] _{r}\cup S.$

3. 3. $dom\left ( E_{r_{1}}\right ) =dom\left ( E_{r}\right ) .$

4. 4. $E_{r_{1}}$ and $E_{q_{1}}$ are equal on $\bigcup \{[ a_{q_{1} }( i) ] _{q_{1}}\mid n<i\}$ (i.e., for every $x\in \bigcup \{[ a_{q_{1}}( i) ] _{q_{1}}\mid n<i\},$ we have that $[ x] _{E_{r_{1}}}=[ x] _{E_{r_{2}}}$ , recall that if $n<i$ then $\left [ a_{q}( i) \right ] _{q}=\left [ a_{r}( i) \right ] _{r}$ ).

Now we find $r_{2}\leq _{n}^{\ast }r_{1}$ (so $r_{2}\leq _{n}^{\ast }r$ ) such that $r_{2}\in D_{r}.$ Let $Z=(dom\left ( r_{1}\right ) \backslash dom\left ( r_{2}\right ) )\cup (\left [ a_{r_{2}}( n) \right ] _{r_{2} }\backslash \left [ a_{r_{1}}( n) \right ] _{r_{1}})$ which is also an element of $\mathcal {U}^{\ast }.$ Let $q_{2}$ be any condition such that:

1. 1. $q_{2}\leq _{n}^{\ast }q_{1}.$

2. 2. $\left [ a_{q_{2}}( n) \right ] _{q_{2}}=\left [ a_{q_{1} }( n) \right ] _{q_{1}}\cup Z.$

3. 3. $dom\left ( E_{q_{2}}\right ) =dom\left ( E_{q_{1}}\right ) .$

4. 4. $E_{q_{2}}$ and $E_{r_{2}}$ are equal on $\bigcup \{\left [ a_{r_{2} }( i) \right ] _{r_{2}}\mid n+1<i\}$ (in the same sense as above).

Since $q_{2}\leq _{n}^{\ast }q_{1}$ and $D_{q}$ is $\leq _{n}^{\ast }$ -open, it follows that $q_{2}\in D_{q}.$ It is clear that $\left \langle q,r,q_{2} ,r_{2}\right \rangle $ has the desired properties.▪

Proof Assume $q=Lim\left ( \left \langle q_{i}\right \rangle _{i\in \omega }\right ) \notin \mathbb {S}( \mathcal {U}) ,$ we will that $r=Lim\left ( \left \langle r_{i}\right \rangle _{i\in \omega }\right ) $ is a condition in $\mathbb {S}( \mathcal {U}) .$ In order to show this, we must first argue that $dom( E_{q_{0}}) =dom( E_{q}) \cup dom( E_{r}) .$ If $n\in dom( E_{q_{0}}) \backslash dom( E_{q}) $ , we then may find $i\in \omega $ such that $n\in dom\left ( q_{i}\right ) \backslash dom\left ( q_{i+1}\right ) $ . Since $\left \langle q_{i},q_{i+1},r_{i},r_{i+1}\right \rangle $ is i-nice then $n\in \left [ a_{r_{i+1}}( i) \right ] _{r_{i+1}}\cup \left [ a_{r_{i+1}}\left ( i+1\right ) \right ] _{r_{i+1}}$ so $n\in dom\left ( E_{r}\right ) .$ Since $dom\left ( E_{q_{0}}\right ) \in \mathcal {U}$ and $\mathcal {U}$ is an ultrafilter, it must be the case that $dom\left ( E_{r}\right ) \in \mathcal {U}.$ ▪

Proof We claim that $q=Lim\left ( \left \langle p_{i}\right \rangle _{i\in \omega }\right ) $ is as desired. We only need to prove that q is really a condition. In order to achieve that, it is enough to prove that $dom\left ( E_{q}\right ) \in \mathcal {U}$ . Since the sequence $\left \langle p_{i}\right \rangle _{i\in \omega }$ is $\leq _{i}^{\ast \ast }$ -decreasing, it follows that $dom\left ( E_{q}\right ) =dom\left ( E_{p_{0}}\right ) $ and we are done.▪

Proof Let $h,g\in 2^{n}$ with $h\neq g$ , we will see that $p[ h] $ and $p\left [ g\right ] $ are incompatible. Let $i<n$ such that $h( i) =0$ and $g( i) =1.$ In this way, we have that $p[ h] \Vdash $ “ $i\notin \dot {x}_{gen}$ ” and $p\left [ g\right ] \Vdash $ “ $i\in \dot {x}_{gen} $ ,” so this two conditions are incompatible.

We now need to prove that the set $\left \{ p[ h] \mid h\in 2^{n}\right \} $ is predense below $p.$ Let $q\leq p,$ take $r\leq q$ such that for every $i<n,$ either $r\Vdash $ “ $i\in \dot {x}_{gen} $ ” or $r\Vdash $ “ $i\notin \dot {x}_{gen} $ .” Define $h:n\longrightarrow 2$ such that $h( i) =1$ if and only if $r\Vdash $ “ $i\in \dot {x}_{gen} $ .” It follows that $r\leq p[ h] .$ ▪

Proof We first show that $\widetilde {D}( p,n) $ is $\leq _{n}^{\ast } $ -dense below $p.$ Let $q\leq _{n}^{\ast }p$ and enumerate $2^{n}=\left \{ h_{i}\mid i<k\right \} $ , we can then recursively find a sequence $\left \langle q_{i}\right \rangle _{i<k+1}$ with the following properties:

1. 1. $q_{0}=q.$

2. 2. $\left \langle q_{i}\right \rangle _{i<k+1}$ is $\leq _{n}^{\ast }$ -decreasing.

3. 3. $q_{i+1}\left [ h_{i}\right ] \in D.$

It is then easy to see that $q_{k+1}\in \widetilde {D}\left ( p,n\right ) .$ Finally, $\widetilde {D}\left ( p,n\right ) $ is $\leq _{n}^{\ast }$ -open, since whenever $r\leq _{n}^{\ast }q$ then $r[ h] \leq q[ h] $ for every $h\in 2^{n}.$ ▪

Proof We will prove that $\mathbb {S}( \mathcal {U}) $ is proper. Let M be a countable elementary submodel of some $H\left ( \kappa \right ) $ and $p\in M.$ Let $\left \{ D_{n}\mid n\in \omega \right \} $ enumerate all open dense subsets of $\mathbb {S}( \mathcal {U}) $ that belong to $M.$ For every $n\in \omega $ let $\widetilde {D}_{n}\left ( p,n\right ) =\left \{ q\leq _{n}^{\ast }p\mid \forall h\in 2^{n}\left ( q[ h] \in D_{n}\right ) \right \} .$ We know that $\widetilde {D}_{n}\left ( p,n\right ) $ is $\leq _{n}^{\ast }$ -open dense. It is also clear that each $\widetilde {D}_{n}\left ( p,n\right ) $ is an element of $M.$ Using Lemma 17, we can construct $T=\left \langle q_{i},r_{i}\right \rangle _{i<\omega }$ a $1$ -fusion sequence with the following properties:

1. 1. $T\subseteq M.$

2. 2. $q_{0},r_{0}\leq p.$

3. 3. $q_{i+1}\in \widetilde {D}_{i}\left ( q_{i},i\right ) $ and $r_{i+1} \in \widetilde {D}_{i}\left ( r_{i},i\right ) $ for every $i\in \omega .$

It is easy to see that if $\overline {p}$ is a fusion of $T,$ then $\overline {p}$ will be an $\left ( M,\mathbb {S}( \mathcal {U}) \right ) $ -generic condition extending $p.$ ▪

Proof Let $p\in \mathbb {S}( \mathcal {U}) $ and $n\in \omega .$ Define $D_{n}=\{q\leq p\mid \exists s\in \omega ^{n+1}(q\Vdash $ “ $\dot {f}\upharpoonright ( n+1) = s$ ” $)\}.$ Clearly $D_{n}$ is an open dense set below $p,$ so by Lemma 26, we get that $\widetilde {D}_{n}\left ( n,p\right ) $ is $\leq _{n}^{\ast }$ -open dense.

Let $q_{0},r_{0}\leq p$ be two incompatible conditions with $E_{q_{0} }=E_{r_{0}}.$ Using the previous remark and Lemma 17, we can construct a $1$ -fusion sequence $T=\left \langle q_{i},r_{i}\right \rangle _{i\in \omega }$ such that $q_{i+1},r_{i+1}\in \widetilde {D}_{i}\left ( i,p\right ) $ for every $i\in \omega .$ By Lemma 20, let $\overline {p}$ be a fusion of $T.$ It follows that $\overline {p}\leq p$ and $\overline {p}$ is $\dot {f}$ -nice.▪

Proof Let $\dot {f}$ be an $\mathbb {S}( \mathcal {U}) $ -name for a new real and $p\in \mathbb {S}( \mathcal {U}) .$ By the previous result, we can find $q\leq p$ that is $\dot {f}$ -nice. For every $n\in \omega ,$ define $S_{n}=\{m\mid \exists h\in 2^{n+1}(q[ h] \Vdash \text{``}\dot {f}( n) =m\text {'' })\}.$ It follows that $S_{n}$ has size at most $2^{n+1}$ and $q\Vdash $ “ $\dot {f}( n) \in S_{n}$ .” This finishes the proof.▪

Proof Let $h:n\longrightarrow 2$ and since $S_{p}\left ( \dot {x}\right ) $ is a Sacks tree, there must be $h ^{\frown }0\subseteq g_{0}^{0}, g_{0}^{1}$ and $h ^{\frown }1\subseteq g_{1}^{0}, g_{1}^{1}$ such that both $\dot {x} [p[g_{0}^{0}]]$ and $\dot {x}[p[g_{0}^{1}]]$ are incompatible and both $\dot {x}[p[g_{1}^{0}]]$ and $\dot {x}[p[g_{1}^{1}]]$ are incompatible. We may also assume there is m such that all $g_{0}^{0}, g_{0}^{1}, g_{1}^{0}$ and $g_{1}^{1}$ have m as its domain. There must be $i,j$ such that $\dot {x}[p[g_{0}^{i}]]$ and $\dot {x}[p[g_{1}^{j}]]$ are incompatible, with out loosing generality, we may assume $i=j=0.$ We define the condition $q\leq p$ with the following properties:

1. 1. $E_{q}=E_{p}\setminus \{\left [ a_{p}( l) \right ] _{p}\mid l\in (n,m)\}.$

2. 2. If $l\in (n,m)$ and $f: 2^{A_{p}\cap a_{p}( l) }\longrightarrow 2$ then the following holds:

   1. (a) If $f( a_{p}( n) ) =0$ then $H_{q} ^{a_{p}( l) }( f) =g_{0}^{0}( l) .$

   2. (b) If $f( a_{p}( n) ) =1$ then $H_{q} ^{a_{p}( l) }( f) =g_{1}^{0}( l) .$

Note that $q\left [ h^{\frown }0\right ] =p[g_{0}^{0}]$ and $q\left [ h^{\frown }1\right ] =p[g_{1}^{0}],$ so the result immediately follows.

The second part of the lemma is proved by applying iteratively the first part.▪

Proof Note that D is $\leq _{n}^{\ast }$ -open by the previous remark, we will see it is also $\leq _{n}^{\ast }$ -dense below $p.$ Let $q\leq _{n}^{\ast }p$ and by the previous lemma, we can find $r\leq _{n}q$ (so $r\leq _{n}^{\ast }q$ ) such that for that $\dot {x}\left [ r\left [ h^{\frown }0\right ] \right ] $ and $\dot {x}\left [ r\left [ h^{\frown }1\right ] \right ] $ are incompatible for every $h:n\longrightarrow 2.$ Since p is $\left ( n-1\right ) $ -separating, r is n-separating.▪

Proof Let $p\in \mathbb {S}( \mathcal {U}) $ be an n-nice condition, we will prove p has a $\omega $ -separating extension. We recursively construct a $1$ -fusion sequence $T=\left \langle q_{i},r_{i}\right \rangle _{i<\omega }$ with the following properties:

1. 1. $q_{0}$ and $r_{0}$ are two incompatible extensions of p with $E_{q_{0}}=E_{r_{0}}.$

2. 2. Both $q_{i},r_{i}$ are i-separating.

This can be done by Lemmas 17 and 36. It is easy to see that any fusion of T will have the desired properties.▪

Proof Let $\dot {x}$ be a $\mathbb {S(\mathcal {U})}$ -name for a new element of $2^{\omega }.$ Fix a condition p that is $\omega $ -separating. Given $n\in \omega , q\leq p$ and $h:n\longrightarrow 2,$ let $I_{q}\left ( h\right ) $ be the union of all the classes $\left [ a_{q}( m ) \right ] _{q}$ with the following properties:

1. 1. $n<m.$

2. 2. For every function $g\in 2^{m}$ with the property that $h\subseteq g$ , we have that $\mathsf {c}_{\mathsf {min}}\left ( \dot {x}\left [ q\left [ g^{\frown }0\right ] \right ] ,\dot {x}\left [ q\left [ g^{\frown }1\right ] \right ] \right ) =1.$

We will now proceed by cases, first assume there are $q\leq p$ and $h\in 2^{<\omega }$ such that $I_{q}\left ( h\right ) \in \mathcal {U}.$ Let $r\leq q[ h] $ such that $dom\left ( E_{r}\right ) =I_{q}\left ( h\right ) $ and every $H_{r}^{l}$ is a constant function for $l\in A_{q} \backslash A_{r}.$ It is then easy to see that $S_{r}\left ( \dot {x}\right ) $ is a $1$ -monochromatic tree and we are done.

Now, we assume that $I_{q}\left ( h\right ) \notin \mathcal {U}$ for every $q\leq p$ and $h\in 2^{<\omega }.$ Given $q\leq p$ and $n<\omega $ let $F_{n}\left ( q\right ) =\{r\leq _{n}^{\ast }q\mid r$ is $\left ( n,0\right ) $ -faithful $\}.$

It is clear that $F_{n+1}\left ( q\right ) $ is $\leq _{n}^{\ast }$ -open below $q,$ we will now prove that it is also $\leq _{n}^{\ast }$ -dense. Let $r\leq _{n}^{\ast }q,$ we will extend r to a $\left ( n,0\right ) $ -faithful condition. Every $\leq _{n}^{\ast }$ -extension of q is n-nice, so by Corollary 37, we may assume that r is $\omega $ -separating. Let $2^{n+1}=\left \{ h_{i}\mid i<k\right \} .$ We know $B=I_{r}\left ( h_{1}\right ) \cup \cdots \cup I_{r}\left ( h_{k}\right ) \in \mathcal {U}^{\ast }$ (since each $I_{r}\left ( h_{i}\right ) \in \mathcal {U}^{\ast }$ by hypothesis) and since B is the union of $E_{r}$ -classes, then there is $m>n+1$ such that $\left [ a_{r}( m ) \right ] _{r}\cap B=\emptyset .$ Since $a_{r}( m ) \notin I_{r}( h_{i}) $ , then for every $i<k$ , there is $g_{i}:m\longrightarrow 2$ extending $h_{i}$ such that $\mathsf {c}_{\mathsf {min}}\left ( \dot {x}\left [ r\left [ g_{i}^{\frown }0\right ] \right ] ,\dot {x}\left [ r\left [ g_{i}^{\frown }1\right ] \right ] \right ) =0$ (recall that r is $\omega $ -separating, so in particular it is m-nice for every $m\in \omega $ ). We now define a condition $r_{1}$ with the following properties:

1. 1. $r_{1}\leq _{n}r.$

2. 2. $dom\left ( E_{r_{1}}\right ) =dom\left ( E_{r}\right ) \backslash \bigcup \left \{ [ a_{r}( l) ] _{r}\mid n<l<m\right \} .$

3. 3. If $m\leq l$ then $ [ a_{r_{1}}( l) ] _{r_{1} }=[ a_{r}( l) ] _{r}.$

4. 4. If $n<l<m$ and $f\in 2^{A_{r_{1}}\cap a_{r}( l) }$ then $H_{r_{1}}^{a_{r}( l) }( f) =g_{i}( l), $ where $i<k$ is (the unique) such that $f\left ( a_{r}( j) \right ) =h_{i}( j) $ for every $j\leq n.$

It is now easy to see that $r_{1}$ is $\left ( n,0\right ) $ -faithful, this finishes the proof of the claim.

We now recursively construct a $1$ -fusion sequence $T=\left \langle q_{i} ,r_{i}\right \rangle _{i<\omega }$ with the following properties:

1. 1. $q_{0}$ and $r_{0}$ are two incompatible extensions of p with $E_{q_{0}}=E_{r_{0}}.$

2. 2. Both $q_{i}$ and $r_{i}$ are $\left ( i,0\right ) $ -faithful.

Let $\overline {p}$ be a fusion of $T.$ It follows that $\overline {p}$ is an extension of p and $S_{\overline {p}}\left ( \dot {x}\right ) $ is a $0$ -monochromatic tree.▪

Proof We will first prove that $\widetilde {D}_{F,\eta }$ is $( F,\eta ) $ -dense. Let p be a condition, enumerate $ {\textstyle \prod \limits _{\delta \in F}} 2^{\eta \left ( \delta \right ) }=\left \{ \sigma _{i}\mid i\leq k\right \} .$ Recursively, we build $\left \{ p_{i}\mid i\leq k\right \} $ such that the following holds:

1. 1. $p_{0}\leq _{F,\eta }p.$

2. 2. $p_{i+1}\leq _{F,\eta }p_{i}$ for $i+1\leq k.$

3. 3. $p_{i}\ast \sigma _{i}\in D.$

This is easy to do. It follows that $p_{k}\in D$ and $p_{k}\leq _{F,\eta }p.$ It is easy to see that $\widetilde {D}_{F,\eta }$ is $( F,\eta ) $ -open.▪

Proof Let $m=\triangle \left ( s,t\right ) ,$ by point 4 of definition 48, we get that $p_{m}$ and $q_{m}$ are incompatible. Since $p_{n+1}\leq p_{m}$ and $q_{n+1}\leq q_{m},$ the result follows.▪

Proof We will first prove point 1. Denote by A the $\sim _{\beta }$ -equivalence class of s (and of z as well, obviously). Let $a=\left ( p_{0} ^{a},\ldots ,p_{n}^{a}\right ) $ and $b=\left ( p_{0}^{b},\ldots ,p_{n}^{b}\right ) $ be two elements of A and $\xi <\alpha .$ We say that $\xi $ is a disagreement point of a and b if the following conditions hold:

1. 1. There is $m\leq n$ such that $\xi \in F_{m+1}\setminus F_{m}.$

2. 2. Denote $\overline {a}=\left ( p_{0}^{a},\ldots ,p_{m-1}^{a}\right ) $ and $\overline {b}=\left ( p_{0}^{b},\ldots ,p_{m-1}^{b}\right ) .$ Either $p_{m} ^{a}=d_{\overline {a}}$ and $p_{m}^{b}=w_{\overline {b}}$ or $p_{m} ^{a}=w_{\overline {a}}$ and $p_{m}^{b}=d_{\overline {b}}.$

Note that if $\xi $ is a disagreement point of a and $b,$ then $\beta \leq \xi $ (this is because $a\sim _{\beta }b$ ). Denote by $dis\left ( a,b\right ) $ the set of all disagreement points of a and $b.$ The following remarks are easy to see:

1. 1. $dis\left ( a,b\right ) $ is a finite set and $\beta \cap dis\left ( a,b\right ) =\emptyset .$

2. 2. $dis\left ( a,b\right ) =\emptyset $ if and only if $a=b.$

3. 3. If $\triangle \left ( a,b\right ) =m$ and $\xi \in F_{m+1}\setminus F_{m},$ then $\xi \in dis\left ( a,b\right ) .$

We will say that a and b are near if $dis\left ( a,b\right ) $ has at most one element. We will now prove the following:

Let $a=\left ( p_{0}^{a},\ldots ,p_{n}^{a}\right ) $ and $b=\left ( p_{0} ^{b},\ldots ,p_{n}^{b}\right ) .$ If $a=b$ , the claim is trivially true, so assume that $a\neq b.$ Let $\xi \in dis\left ( a,b\right ) $ and $m\in \omega $ such that $\xi \in F_{m+1}\setminus F_{m}.$ We claim that $a\sim _{\xi }b$ (in other words, $p_{i}^{a}\upharpoonright \xi =p_{i}^{b}\upharpoonright \xi $ for every $i\leq n$ ).

1. 1. If $i<m,$ then $p_{i}^{a}=p_{i}^{b}$ (so $p_{i}^{a}\upharpoonright \xi =p_{i}^{b}\upharpoonright \xi $ ).

2. 2. If $i=m,$ then $p_{i}^{a}\upharpoonright \xi =p_{i}^{b}\upharpoonright \xi $ by point 5(a) of Definition 48.

3. 3. If $i>m,$ then $p_{i}^{a}\upharpoonright \xi =p_{i}^{b}\upharpoonright \xi $ because T is fusion tree and $\xi $ is the only point of disagreement between a and $b.$

We will provide more details for the last point. We proceed by induction over $i.$ Assume that $p_{i}^{a}\upharpoonright \xi =p_{i}^{b}\upharpoonright \xi $ for $m\leq i<n,$ we will prove that $p_{i+1}^{a}\upharpoonright \xi =p_{i+1} ^{b}\upharpoonright \xi .$ In case $F_{i+1}=F_{i},$ the conclusion follows by point 5(b) of Definition 48. Now assume that $F_{i+1}\neq F_{i}$ and let $\delta \in F_{i+1}\setminus F_{i}.$ Since $\delta $ is not a disagreement point, by point 5(a) of Definition 48, we conclude that $p_{i+1}^{a}\upharpoonright \xi =p_{i+1}^{b}\upharpoonright \xi .$

In particular, we get that $p_{n}^{a}\upharpoonright \xi =p_{n}^{b} \upharpoonright \xi .$ By point 5 of Definition 48, we get that $d_{a}\upharpoonright \xi =d_{b}\upharpoonright \xi $ and $w_{a} \upharpoonright \xi =w_{b}\upharpoonright \xi .$ Since $\beta \leq \xi ,$ we conclude that $d_{a}\upharpoonright \beta =d_{b}\upharpoonright \beta $ and $w_{a} \upharpoonright \beta =w_{b}\upharpoonright \beta .$ This finishes the proof of the claim.

Let $a=\left ( p_{0}^{a},\ldots ,p_{n}^{a}\right ) , b=\left ( p_{0} ^{b},\ldots ,p_{n}^{b}\right ) , m=\triangle \left ( a,b\right ) $ , and $\xi \in F_{m+1}\setminus F_{m}.$ Define $c=\left ( p_{0}^{c},\ldots ,p_{n}^{c}\right ) $ as follows:

1. 1. $p_{i}^{c}=p_{i}^{a}$ for $i<m$ (so $p_{i}^{c}=p_{i}^{b}$ ).

2. 2. $p_{m}^{c}=p_{m}^{b}.$

3. 3. Let $i\geq m,$ denote $\overline {a}=\left ( p_{0}^{a},\ldots ,p_{i} ^{a}\right ) $ and $\overline {c}=\left ( p_{0}^{c},\ldots ,p_{i}^{c}\right ) .$ Let $p_{i+1}^{c}=d_{\overline {c}}$ if $p_{i+1}^{a}=d_{\overline {a}}$ and $p_{i+1}^{c}=w_{\overline {c}}$ if $p_{i+1}^{a}=w_{\overline {a}}.$

Note that c is in the tree T since we are always taking the successor of a node in $T.$ It follows by the construction that $\triangle \left ( a,c\right ) =m$ and by point 5 of Definition 48, we get that $p_{i} ^{c}\upharpoonright \xi =p_{i}^{a}\upharpoonright \xi $ for all $i\leq n,$ hence $c\in A.$ It is clear that a and c are near and $\left \vert dis\left ( c,b\right ) \right \vert =\left \vert dis\left ( a,b\right ) \right \vert -1.$ This finishes the proof of the claim.

We can now finish the proof of the first point of the lemma. Recall that $s\sim _{\beta }z$ and we wanted to prove that $d_{s}\upharpoonright \beta =d_{z}\upharpoonright \beta $ and $w_{s}\upharpoonright \beta =w_{z} \upharpoonright \beta $ . Now, by claim 53, we can find a sequence $\left ( a_{0},\ldots ,a_{k}\right ) $ such that for every $i\leq k,$ the following conditions hold:

1. 1. $a_{i}\in A.$

2. 2. $a_{0}=s$ and $a_{k}=z.$

3. 3. $a_{i+1}$ is near of $a_{i}.$

By Claim 52, we conclude that $d_{s}\upharpoonright \beta =d_{z}\upharpoonright \beta $ and $w_{s}\upharpoonright \beta =w_{z} \upharpoonright \beta $ .

The second point of the lemma follows by the first one, we will now prove the third point. Assume that $F_{n+1}^{\prime }\setminus F_{n}^{\prime } \neq \emptyset ,$ we want to prove that $d_{s}\upharpoonright \beta \neq w_{s}\upharpoonright \beta .$ Let $\gamma \in F_{n+1}^{\prime }\setminus F_{n}^{\prime }.$ Since $F_{n+1}^{\prime }=F_{n+1}\cap \beta $ and $F_{n}^{\prime }=F_{n}\cap \beta ,$ it follows that $\gamma <\beta $ . By point 4 of definition 48, we know that $d_{s}$ and $w_{s}$ differ in the $\gamma $ component, which entails that $d_{s}\upharpoonright \beta \neq w_{s} \upharpoonright \beta .$

We now prove point 4 of the lemma. Assume that $F_{n+1}^{\prime }\setminus F_{n}^{\prime }=\emptyset ,$ we want to prove that $d_{s}\upharpoonright \beta =w_{s}\upharpoonright \beta .$ If $F_{n+1}\setminus F_{n}=\emptyset ,$ it follows by point 3 of Definition 48 that $d_{s}=w_{s},$ so $d_{s}\upharpoonright \beta =w_{s}\upharpoonright \beta .$ Assume now that $F_{n+1}\setminus F_{n}\neq \emptyset .$ Let $\gamma \in F_{n+1}\setminus F_{n}.$ Since $F_{n+1}^{\prime }\setminus F_{n}^{\prime }=\emptyset ,$ it must be the case that $\beta \leq \gamma .$ By point 5 of Definition 48 we know that $d_{s}\upharpoonright \gamma =w_{s}\upharpoonright \gamma ,$ which implies that $d_{s}\upharpoonright \beta =w_{s}\upharpoonright \beta .$ ▪

Proof We already noted that $L^{\prime }$ is suitable for $\left ( M,\beta \right ) .$ Let $S=T\upharpoonright \beta .$ For this proof, we adopt the following convention: if $s=\left \langle q_{0},\ldots ,q_{n}\right \rangle \in T,$ define $s^{\prime }=\left \langle q_{0}\upharpoonright \beta ,\ldots ,q_{n}\upharpoonright \beta \right \rangle $ (we are not assuming this convention outside this proof, unless when we mention it explicitly). Note that $S_{n}=\left \{ s^{\prime }\mid s\in T_{n}\right \} $ for every $n\in \omega .$ We will prove that S is a $\beta $ -fusion tree. The first point from Definition 48 is clear.

Note that for every $s\in S$ there may be many $t\in T$ such that $t^{\prime }=s.$ This is where Lemma 51 comes into play. It is easy to see that $suc_{S}( s) =\left \{ x\upharpoonright \beta \mid \exists t\in T\left ( t^{\prime }=s\wedge x\in suc_{T}\left ( t\right ) \right ) \right \} .$ Point 2 from Definition 48 follows by point 1 of Lemma 51. Point 3 of the definition of fusion tree follows by point 4 of Lemma 51. Point 4 follows from the definition of $T\upharpoonright \beta $ and Lemma 51.